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3 years ago

A box has a mass of 6.0 kg. What is its weight on the mars, where the acceleration due to

Physics

1 answer:

Dvinal [7]3 years ago

3 0

Answer:

The weight of the box is 22.38 N.

Explanation:

The weight (W) of the mass is given by:

$W = mg$

Where:

m: is the mass of the box = 6.0 kg

$g_{m}$ : is the acceleration due to gravity of mars

Since the acceleration due to the gravity of mars is 0.38 times that on Earth we have:

$g_{m} = 0.38g_{E} = 0.38*9.81 m/s^{2} = 3.73 m/s^{2}$

Hence, the weight of the mass is:

$W = mg = 6.0 kg*3.73 m/s^{2} = 22.38 N$

Therefore, the weight of the box is 22.38 N.

I hope it helps you!

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A small remote-control car with a mass of 1.65 kg moves at a constant speed of v = 12.0 m/s in a vertical circle inside a hollow

Reil [10]

Answer:

$N=63.69N$

Explanation:

The normal force exerted on the car by the walls of the cylinder at the bottom of the vertical circle will be such that when substracted to the weight it must give the centripetal force, since at that point on the vertical $F_{cp}=N-W=N-mg$

We also know that the equation for the centripetal force is:

$F_{cp}=ma_{cp}=\frac{mv^2}{r}$

Mixing both equations we get:

$N-mg=\frac{mv^2}{r}$

$N=mg+\frac{mv^2}{r}$

Which for our values means:

$N=(1.65Kg)(9.8m/s^2)+\frac{(1.65Kg)(12m/s)^2}{(5m)}=63.69N$

8 0

3 years ago

A marble, rolling with speed of 20m/sec rolls off the edge of the table that is 180m high (g=10m/sec2), find time taken to drop

natali 33 [55]

Answer:

<em>Choice: c. 6sec</em>

Explanation:

<u>Horizontal Launch </u>

When an object is thrown horizontally with a speed (v) from a height (h), it describes a curved path ruled by gravity until it finally hits the ground.

The horizontal component of the velocity is always constant because no acceleration exists in that direction, thus:

$v_x=v$

The vertical component of the velocity changes in time because gravity makes the object fall at increasing speed given by:

$v_y=g.t$

Where $g=10 m/s^2$

To calculate the time the object takes to hit the ground, we use the same formula as for free-fall, since the time does not depend on the initial speed:

$\displaystyle t=\sqrt{\frac{2h}{g}}$